See the attached file.
A car has a mass of 1.4 tonnes. The distance between the front and rear wheel axles is 3.6m and the centre of mass is positioned 1.5m behind the front wheel centre and 0.8m above the road level. These details are attached in Figure Q1.
The car, initially traveling at 12.6km/h, is accelerated up a straight road inclined at 12 degrees to the horizontal so that it covers a distance of 1.2km in a time of 40s.
Determine the following:
(a) the traction force
(b) the wheel reactions
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a) First lets calculate the acceleration we have vo = 12.6 km/hour a distance of 1.2 km traversed in 40 seconds
x = vot + 0.5 * a * t^2 thus x -vot = 0.5 * a * t^2 and a = 2(x-vot)/t^2 so a = 2(1200 m - 12,600 m/3600secx 40 sec) / (40sec)^2 = 1.35 m/sec^2. This is the net acceleration
The net force is then 1.4 tonne x 2000Kg/tonne x 1.35 m/sec^2 = 3710 N
Here I have assumed a metric tonne.
Now the traction force must also overcome the component of the weight of the car that is in the direction of this acceleration. The angle of the grade is 12 degrees The weight ...
The solution recognises that there are a number of other things that must be calculated before one can find the traction and the reaction of the wheels for the described car to accelerate up an inclined road and it walks the reader through each of them with written explanation and clearly calculated steps.